Step 8Why does this collector design work so well?
There is a property of any thermal system called thermal conductance that indicates how much heat (power) can be transfered from point 'a' to point 'b' for a given temperature differential. The formula is:
Thermal Conductance = K * A / L
where:
K = thermal conductivity (a physical property of the material)
A = cross-sectional area through which heat must travel
L = distance heat must travel (the distance from 'a' to 'b').
Lets calculate the thermal conductance of a typical flat panel collector.
Assume the panel is 2'x8' with 4 copper tubes running lengthwise and fins sticking out 3" on either side of every tube (6" per tube x 4 tubes fills our 2' width). Suppose the fins are 1mm thick and also made of copper. When the fins heat up, the cross sectional area through which this heat must be conducted to reach the tubes is 1mm * 8 ft * 8 fins = 2438 mm2. The average distance the heat must be conducted is 1/2 the fin width or 1.5" = 38 mm. The conductivity of copper is about 0.4 W/mm/degreeC.
Therefore the thermal conductance from the collector surface to the water is 0.4 W/mm/degreeC * 2438 mm2 / 38 mm = 25W/degreeC. In other words, a 1 degreeC temperature difference between the water and the fin will result in 25W of heat transfer. But the panel is receiving something on the order of 1400 W of incoming power from sunlight. To transfer all that power to the water by conduction alone the fins would have to heat up to 56 degrees C higher than the water temperature.
Now repeat the calculation for the corrugated plastic panel.
The cross sectional area through which heat must be conducted is the receiving area of the panel itself (2' * 8' = 1486448 mm2). The distance the heat must travel to reach the water is just the thickness of the plastic wall or about 0.3mm. The conductivity of plastic is about 0.0001 W/mm/degreeC. Note that it is over 1000 times lower than copper which makes sense since plastic is general thought of as an insulator, not a conductor.
Therefore the thermal conductance of the system is 0.0001 W/mm/degreeC * 1486448 mm2 / 0.3 mm = 495 W/degreeC. In other words, a 1 degreeC temperature difference between the water and the collector surface will result in 495 W of heat transfer into the water. To transfer 1400W, the panel surface only needs to heat up about 3 degreesC hotter than the water.
Of course in practice, not all of that 1400W goes into the water. The conductance from the collector surface to the water is in parallel with another conductance from the collector surface to the outside air. The relative values of those two conductances determines how much heat goes where (Aside: this is analogous to current in an electrical circuit with two resistors in parallel.)
Conclusion
You can see that in spite of the much lower thermal conductivity of plastic, using a corrugated plastic sheet as a collector achieves 20 times higher conductance between the collector surface and the water when compared to a typical tube-and-fin design.
If a corrugated collector could be made from copper, the results would be even better, but not very much better, for reasons I won't go into because I can already feel everyone's eyes glazing over.
Thanks for reading. For information on this and other projects of mine see my website IWillTry.org.
| « Previous Step | Download PDFView All Steps | Next Step » |
16
comments
|
Add Comment
|
iwilltry (author)
says:
iwilltry (author)
says:
iwilltry (author)
says:
iwilltry (author)
says:
iwilltry (author)
says:
![]() |
Add Comment
|













































